JEE Mains · Chemistry · STD 11 - 6.2. Equilibrium - II (icon Equilibrium)
The solubility product of a sparingly soluble salt \(A _{2} X _{3}\) is \(1.1 \times 10^{-23}\). If specific conductance of the solution is \(3 \times 10^{-5} \,S\, m ^{-1}\), the limiting molar conductivity of the solution is \(x \times 10^{-3} \,S\, m ^{2}\, mol ^{-1}\). The value of \(x\) is ...
- A \(30\)
- B \(54\)
- C \(3\)
- D \(90\)
Answer & Solution
Correct Answer
(C) \(3\)
Step-by-step Solution
Detailed explanation
\(A _{2} X _{3( s )} \rightleftharpoons 2 A _{( aq )}^{+3}+3 X _{( aq )}^{-2}\) solubility \(= sM\quad 2 s \quad 3 s\) \((2 s )^{2}(3 s )^{3}=1.1 \times 10^{-23}\) \(108 s ^{5}=1.1 \times 10^{-23}\) \(s \simeq 10^{-5} M =10^{-5} \,\frac{ mol }{ L }=0.01 \,\frac{ mol }{ m ^{3}}\)…
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